So from #arccos(x) + arccos(2x) = 60º# take the cosine from both sides
#cos(arccos(x)+arccos(2x)) = 1/2#
Expand the cosine sum, and use #cos(arccos(theta)) = theta# to rewrite it
#x*2x - sin(arccos(x))sin(arccos(2x)) = 1/2#
Use the pythagorean identity #sin^2(x) = 1 - cos^2(x)# to rewrite it. Knowing that the sine is always positive on the arccosine range.
#x*2x - sqrt(1-x^2)*sqrt(1-4x^2)=1/2#
#2x^2 - sqrt((1-x^2)(1-4x^2)) = 1/2#
Isolate the root, then square both sides
#-sqrt((1-x^2)(1-4x^2)) = 1/2 - 2x^2#
#(1-x^2)(1-4x^2) = 1/4 - 2x^2 + 4x^4#
Simplify,
#1 - x^2 -4x^2 + 4x^4 = 1/4 - 2x^2 + 4x^4#
#1 -5x^2 = 1/4 -2x^2#
#1 - 1/4 = 5x^2 -2x^2#
#3/4 = 3x^2#
#1/4 = x^2#
Taking the square root
#x = +- 1/2#
